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For the hypothesis test \(H_{0}: \mu=5\) against \(H_{1}: \mu<5\) and variance known, calculate the \(P\) -value for each of the following test statistics. (a) \(z_{0}=2.05\) (b) \(z_{0}=-1.84\) (c) \(z_{0}=0.4\)

Short Answer

Expert verified
(a) P-value = 0.9798; (b) P-value = 0.0329; (c) P-value = 0.6554.

Step by step solution

01

Understanding the Hypotheses

We have a null hypothesis \( H_0: \mu = 5 \) and an alternative hypothesis \( H_1: \mu < 5 \). Since the alternative hypothesis is one-sided, we are conducting a left-tailed test.
02

Determine the Nature of Each Test Statistic

For Part (a) with \( z_0 = 2.05 \), the test statistic is greater than zero. For Part (b), \( z_0 = -1.84 \) is negative, which indicates a result possibly supporting \( H_1 \). For Part (c), \( z_0 = 0.4 \) is positive, similarly indicating results more aligned with \( H_0 \).
03

Calculating the P-value for Part (a)

Since \( z_0 = 2.05 \) is positive, calculate the \( P \)-value by finding \( P(Z < 2.05) \) using the standard normal distribution table.\\( P(Z < 2.05) = 0.9798 \). Since the alternative hypothesis is left-tailed, the \( P \)-value is \( 1 - P(Z < 2.05) = 0.9798 \).
04

Calculating the P-value for Part (b)

With \( z_0 = -1.84 \), find \( P(Z < -1.84) \) using the standard normal distribution table.\\( P(Z < -1.84) = 0.0329 \). Since this is a left-tailed test, the \( P \)-value is \( 0.0329 \).
05

Calculating the P-value for Part (c)

For \( z_0 = 0.4 \), determine \( P(Z < 0.4) \) using the standard normal distribution table.\\( P(Z < 0.4) = 0.6554 \). For a left-tailed test, the \( P \)-value remains \( 0.6554 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Z-Test
A z-test is a statistical method used to determine if there is a significant difference between the means of two groups. It is particularly useful when the population variance is known and the sample size is large (usually over 30). In our example, we have the null hypothesis, \( H_0: \mu = 5 \), which means we believe the population mean is 5. The alternative hypothesis, \( H_1: \mu < 5 \), suggests that the population mean is less than 5.
This particular hypothesis test is a one-tailed test since we're only interested in whether the mean is less than the hypothesized value. The z-test involves calculating a test statistic, z, which tells us how far and in what direction our sample mean deviates from the hypothesized population mean, in unit of the standard error.\( z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \) is the formula for the z-test statistic, where \( \bar{x} \) is the sample mean, \( \mu \) is the hypothesized population mean, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
It's important to first understand whether you have a positive or negative z-value, as this determines whether your sample mean is larger or smaller than the population mean. In our exercise, we have to calculate the z-value and compare it to a critical value from the standard normal distribution to interpret the results effectively.
Exploring P-value Calculation
The P-value is a key concept in hypothesis testing. It represents the probability of observing test results as extreme as the results actually observed, under the assumption that the null hypothesis is true. In simpler terms, it's the evidence against a null hypothesis.
In the standard normal distribution, we interpret the z-value using the P-value. A low P-value, typically less than 0.05, indicates strong evidence against the null hypothesis, leading us to reject it. Conversely, a high P-value suggests that the sample provides insufficient evidence to decide against the null hypothesis.
For example, if the P-value from our calculated z-test is 0.0329, it indicates that there’s only a 3.29% probability that the population mean is 5 under the given conditions. This would provide strong evidence against the null hypothesis. Calculating P-values involves looking at the standard normal distribution table which gives us a cumulative probability. For one-tailed tests, this is straightforward as you focus on one end of the distribution curve. Hence, understanding how to find and interpret P-values is essential for determining statistical significance in the context of hypothesis testing.
One-tailed Test Clarified
When conducting hypothesis tests, there are one-tailed and two-tailed tests. In a one-tailed test, the alternative hypothesis specifies the direction of the effect. It tests either if the parameter is greater than or less than a certain value—not both.
For our exercise, because the alternative hypothesis is written as \( H_1: \mu < 5 \), we are interested if the mean is significantly less than 5. This means the test is one-tailed and more specifically, it is a left-tailed test. In contrast, a right-tailed test would explore if the mean is greater than a specified value.
Understanding the type of test is crucial as it affects how we calculate and interpret the P-values. With a one-tailed test, you are focusing on one end of the distribution, making the threshold for rejecting the null hypothesis different than if it was two-tailed. This is because, in a one-tailed test, all the significance or rejection is tested on one side of the distribution curve instead of being split between two.

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Most popular questions from this chapter

A 1992 article in the Journal of the American Medical Association ("A Critical Appraisal of 98.6 Degrees \(\mathrm{F}\), the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich") reported body temperature, gender, and heart rate for a number of subjects. The body temperatures for 25 female subjects follow: \(97.8,97.2,97.4,97.6,\) 97.8,97.9,98.0,98.0,98.0,98.1,98.2,98.3,98.3,98.4,98.4 \(98.4,98.5,98.6,98.6,98.7,98.8,98.8,98.9,98.9,\) and \(99.0 .\) (a) Test the hypothesis \(H_{0}: \mu=98.6\) versus \(H_{1}: \mu \neq 98.6\), using \(\alpha=0.05 .\) Find the \(P\) -value. (b) Check the assumption that female body temperature is normally distributed. (c) Compute the power of the test if the true mean female body temperature is as low as 98.0 . (d) What sample size would be required to detect a true mean female body temperature as low as 98.2 if we wanted the power of the test to be at least \(0.9 ?\) (e) Explain how the question in part (a) could be answered by constructing a two-sided confidence interval on the mean female body temperature.

Suppose that 500 parts are tested in manufacturing and 10 are rejected. (a) Test the hypothesis \(H_{0}: p=0.03\) against \(H_{1}: p<0.03\) at \(\alpha=0.05 .\) Find the \(P\) -value. (b) Explain how the question in part (a) could be answered by constructing a \(95 \%\) one-sided confidence interval for \(p\).

For the hypothesis test \(H_{0}: \mu=10\) against \(H_{1}: \mu>10\) with variance unknown and \(n=15,\) approximate the \(P\) -value for each of the following test statistics. (a) \(t_{0}=2.05\) (b) \(t_{0}=-1.84\) (c) \(t_{0}=0.4\)

A hypothesis will be used to test that a population mean equals 5 against the alternative that the population mean is less than 5 with known variance \(\sigma\). What is the criti- cal value for the test statistic \(Z_{0}\) for the following significance levels? (a) \(\alpha=0.01\) and \(n=20\) (b) \(\alpha=0.05\) and \(n=12\) (c) \(\alpha=0.10\) and \(n=15\)

A hypothesis will be used to test that a population mean equals 7 against the alternative that the population mean does not equal 7 with unknown variance \(\sigma\). What are the critical values for the test statistic \(T_{0}\) for the following significance levels and sample sizes? (a) \(\alpha=0.01\) and \(n=20\) (b) \(\alpha=0.05\) and \(n=12\) (c) \(\alpha=0.10\) and \(n=15\)

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